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andrew11 [14]
2 years ago
10

The headlights are shining on a truck travelling at 100 km/h. The speed of the light from the headlights relative to the road wi

ll be:
A) c

B) c + 100 km/h

C) c – 100 km/h

D) depends on the temperature, but faster than the speed if the truck was not moving.

E) faster than if the truck was not moving, but impossible to calculate with the given information
Physics
2 answers:
Lady_Fox [76]2 years ago
7 0

Answer:

A) c

Explanation:

Speed of light is always constant irrespective of the wave source's motion and the observer's inertial frame of reference. So, no matter how fast the car is moving the speed of light will always be constant. The speed of light in air is around 299704644.54 m/s. The meter is also defined by the speed of light as 1 meter is the distance travelled by light in 1/299792458 second.

jek_recluse [69]2 years ago
3 0
A, the light generated from the headlights travels at a fixed speed relative to the road
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A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the disk is allowed to fall and th
Tju [1.3M]

Answer:

a = \frac{2}{3}g

T = \frac{mg}{3}

Explanation:

As the disc is unrolling from the thread then at any moment of the time

We have force equation as

mg - T = ma

also by torque equation we can say

TR = I\alpha

TR = \frac{1}{2}mR^2(\frac{a}{R})

T = \frac{1}{2}ma

Now we have

mg - \frac{1}{2}ma = ma

mg = \frac{3}{2}ma

a = \frac{2}{3}g

Also from above equation the tension force in the string is

T = \frac{1}{2}ma

T = \frac{mg}{3}

7 0
2 years ago
The point on the graph that lies on the y-axis (vertical axis) is called the y-intercept. What does the y-intercept tell you abo
jekas [21]

Answer:

The starting position of the runner.

Explanation:

When you look at the graph, you can see that the first point on the graph is twenty on the y-axis.

The runner starts at twenty, and ends at thirty.

Therefore, the runner starts at twenty on the y-axis, so it's the starting position of the runner.

7 0
2 years ago
A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal
Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration (g=9.8 m/s^2) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

v=18.9 m/s

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

v_y^2 - u_y^2 = 2ah (1)

where

u_y = 16.4 m/s is the initial vertical velocity

v_y is the vertical velocity at height h

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, v_y = 0, so we can use the equation to find the maximum height

h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m

So, at half of the maximum height,

h = \frac{13.7}{2}=6.9 m

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s

And so, the speed of the stone at half of the maximum height is

v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

v_1 = 18.9 m/s

while the speed at half of the maximum height (part b) is

v_2 = 22.2 m/s

So the difference is

\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s

And so, in percentage,

\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0
2 years ago
On a warm summer day (31 ∘c), it takes 4.60 s for an echo to return from a cliff across a lake. on a winter day, it takes 5.00 s
xenn [34]
The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.

First of all, we need to calculate the speed of sound at temperature of T=31^{\circ}C:
v=(331+0.60 T) m/s = (331+0.6 \cdot 31) m/s = 349.6 m/s

The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
v= \frac{2L}{t}
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
L= \frac{vt}{2}= \frac{(349.6 m/s)/4.60 s)}{2}=  804.1 m
6 0
2 years ago
You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.
Andru [333]

Answer:

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Explanation:

Hi there!

The equations of height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity of the ball at time t.

Placing the origin at the throwing point, y0 = 0.

Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.

v = v0 + g · t

6.00 m/s = 12.0 m/s -9.81 m/s² · t

(6.00 - 12.0)m/s / -9.81 m/s² = t

t = 0.612 s

Now, let´s calculate the height of the baseball at that time:

y = y0 + v0 · t + 1/2 · g · t²     (y0 = 0)

y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²

y = 5.51 m

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Have a nice day!

4 0
2 years ago
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